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# HW01 - β x Β 2 β x 2 β x = x(4-x Β(2 β x 2 2 β x...

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MATH 132 HW 1 Answer Section 2.2 [18]. Find the limits lim h 0 5 h + 4 - 2 h Solution: 5 · 0+4 - 2 0 = 0 0 is not defined. 5 h + 4 - 2 h = 5 h + 4 - 2 h · 5 h + 4 + 2 5 h + 4 + 2 = ( 5 h + 4 ) 2 - 2 2 h ( 5 h + 4 + 2 ) = 5 h h ( 5 h + 4 + 2 ) = 5 5 h + 4 + 2 -→ 5 4 as h 0 [24]. Find the limits lim t →- 1 t 2 + 3 t + 2 t 2 - t - 2 Solution: ( - 1) 2 +3( - 1)+2 ( - 1) 2 - ( - 1) - 2 = 0 0 is not defined. t 2 + 3 t + 2 t 2 - t - 2 = ( t + 2) ( t + 1) ( t - 2) ( t + 1) = ( t + 2) ( t - 2) -→ 1 - 3 as t → - 1 [30]. Find the limits lim x 4 4 x - x 2 2 - x Solution: 4 · 4 - 4 2 2 - 4 = 0 0 is not defined. 4
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Unformatted text preview: β x Β· 2 + β x 2 + β x = x (4-x ) Β· (2 + β x ) 2 2-( β x ) 2 = x (4-x ) Β· (2 + β x ) 4-x = x ( 2 + β x )-β 4 Β± 2 + β 4 Β² = 16 as x β 4 Ex (2) : lim x β sin 2 x 5 x = lim x β sin 2 x 2 x Β· 2 5 = lim x β sin 2 x 2 x Β· lim x β 2 5 = 1 Β· 2 5 = 2 5 1...
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